If a variable line drawn through the intersection of the lines {x \over 3} + {y \over 4} = 1 and {x \over 4} + {y \over 3} = 1, meets the coordinate axes at A and B, (A
e B), then the locus of the midpoint of AB is :

Question

If a variable line drawn through the intersection of the lines {x \over 3} + {y \over 4} = 1 and {x \over 4} + {y \over 3} = 1, meets the coordinate axes at A and B, (A 
e B), then the locus of the midpoint of AB is :

Accepted Answer

Answer: C. 7xy = 6(x + y) L1 : 4x + 3y - 12 = 0 L2 : 3x + 4y - 12 = 0 Equation of line passing through the intersection of these two lines L1 and L2 is L1 + \lambda L2 = 0 \Rightarrow \,\,\,(4x + 3y - 12) + \lambda (3x + 4y - 12) = 0 \Rightarrow \,\,\, x(4 + 3\lambda ) + y(3 + 4\lambda ) - 12(1 + \lambda ) = 0 this line meets x coordinate at point A and y coordinate at point B. 	herefore\,\,\, Point A = \left( {{{12\left( {1 + \lambda } ight)} \over {4 + 3\lambda }},0} ight) and Point B = \left( {0,\,\,{{12\left( {1 + \lambda } ight)} \over {3 + 4\lambda }}} ight) Let coordinate of midpoint of line AB is (h, k). 	herefore\,\,\, h = {{6\left( {1 + \lambda } ight)} \over {4 + 3\lambda }} . . . . . (1) and k = {{6\left( {1 + \lambda } ight)} \over {3 + 4\lambda }} . . . . (2) Eliminate \lambda from (1) and (2), then we get 6(h + k) = 7 hk 	herefore\,\,\, Locus of midpoint of line AB is , 6(x + y) = 7xy

If a variable line drawn through the intersection of the lines $${x \over 3} + {y \over 4} = 1$$ and $${x \over 4} + {y \over 3} = 1,$$ meets the coordinate axes at A and B, (A $$ \ne $$ B), then the locus of the midpoint of AB is :

Answer

Explanation

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